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Help on Geometry Problem - OG 13 - PS#75

This topic has 4 expert replies and 3 member replies
amirhakimi Senior | Next Rank: 100 Posts
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Help on Geometry Problem - OG 13 - PS#75

Post Thu Oct 31, 2013 6:04 am
I can't understand this problem:

In the figure above, triangle ABC is equilateral, and point P is equidistant from vertices A, B, and C. If triangle ABC is rotated clockwise about point P, what is the minimum number of degrees the triangle must be rotated so that point B will be in the position where point A is now?
(A) 60
(B) 120
(C) 180
(D) 240
(E) 270

I already read the explanation but still no clues to handle this question type. Please help me understand this solution.

Since ABC is equilateral, the measure of
Then rotating this figure clockwise about point P through an angle of 120° will produce the figure shown below.

In this figure, point B is in the position where points was in the original figure. The triangle was rotated clockwise about point P through 120° + 120° = 240°.
The correct answer is D.



Last edited by amirhakimi on Thu Oct 31, 2013 7:43 am; edited 1 time in total

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mcdesty Master | Next Rank: 500 Posts
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Post Thu Jul 10, 2014 9:16 am
Half a circle(180 degrees) plus the 60 degrees for one angle of the equilateral triangle. See attachment.
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Post Tue Jun 23, 2015 3:04 pm
amirhakimi wrote:
I can't understand this problem:

In the figure above, triangle ABC is equilateral, and point P is equidistant from vertices A, B, and C. If triangle ABC is rotated clockwise about point P, what is the minimum number of degrees the triangle must be rotated so that point B will be in the position where point A is now?
(A) 60
(B) 120
(C) 180
(D) 240
(E) 270

I already read the explanation but still no clues to handle this question type. Please help me understand this solution.

Since ABC is equilateral, the measure of
Then rotating this figure clockwise about point P through an angle of 120° will produce the figure shown below.

In this figure, point B is in the position where points was in the original figure. The triangle was rotated clockwise about point P through 120° + 120° = 240°.
The correct answer is D.
Solution:

To more easily solve this problem we can inscribe triangle ABC in a circle. Remember that a circle has a total of 360 degrees. So, if we were to rotate point B around the ENTIRE circle, it would rotate 360 degrees, returning to its original position. We also know that the inscribed equilateral triangle breaks the circle into 3 arcs of equal length. In other words, arc AB = arc BC = arc CA. We can take this one step further and say that each arc equals 360/3 = 120 degrees. So the degree measurement from B to C is 120 degrees, from C to A is 120 degrees and from A to B is 120 degrees. Let's see what happens when we rotate the triangle clockwise. Let’s first rotate the triangle so that point B ends up where point C currently is. In order to get B to the position where C is, we have to rotate the triangle 120 degrees. Finally, we must rotate the triangle one more time so that point B is where point A initially was. Notice that to get point B to where point A initially was, we have to again rotate the triangle 120 degrees. Thus, in total, we have rotated the triangle 240 degrees to get point B to the position where point A originally was.

Answer: D

ttp://postimg.org/image/mswepes9t/" target="_blank">

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mcdesty Master | Next Rank: 500 Posts
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Post Thu Jul 10, 2014 9:16 am
Half a circle(180 degrees) plus the 60 degrees for one angle of the equilateral triangle. See attachment.
Attachments

This post contains an attachment. You must be logged in to download/view this file. Please login or register as a user.

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Post Tue Jun 23, 2015 3:04 pm
amirhakimi wrote:
I can't understand this problem:

In the figure above, triangle ABC is equilateral, and point P is equidistant from vertices A, B, and C. If triangle ABC is rotated clockwise about point P, what is the minimum number of degrees the triangle must be rotated so that point B will be in the position where point A is now?
(A) 60
(B) 120
(C) 180
(D) 240
(E) 270

I already read the explanation but still no clues to handle this question type. Please help me understand this solution.

Since ABC is equilateral, the measure of
Then rotating this figure clockwise about point P through an angle of 120° will produce the figure shown below.

In this figure, point B is in the position where points was in the original figure. The triangle was rotated clockwise about point P through 120° + 120° = 240°.
The correct answer is D.
Solution:

To more easily solve this problem we can inscribe triangle ABC in a circle. Remember that a circle has a total of 360 degrees. So, if we were to rotate point B around the ENTIRE circle, it would rotate 360 degrees, returning to its original position. We also know that the inscribed equilateral triangle breaks the circle into 3 arcs of equal length. In other words, arc AB = arc BC = arc CA. We can take this one step further and say that each arc equals 360/3 = 120 degrees. So the degree measurement from B to C is 120 degrees, from C to A is 120 degrees and from A to B is 120 degrees. Let's see what happens when we rotate the triangle clockwise. Let’s first rotate the triangle so that point B ends up where point C currently is. In order to get B to the position where C is, we have to rotate the triangle 120 degrees. Finally, we must rotate the triangle one more time so that point B is where point A initially was. Notice that to get point B to where point A initially was, we have to again rotate the triangle 120 degrees. Thus, in total, we have rotated the triangle 240 degrees to get point B to the position where point A originally was.

Answer: D

ttp://postimg.org/image/mswepes9t/" target="_blank">

_________________

Scott Woodbury Stewart Founder & CEO
GMAT Quant Self-Study Course - 500+ lessons 3000+ practice problems 800+ HD solutions
5-Day Free Trial 5-DAY FREE, FULL-ACCESS TRIAL TTP QUANT

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Matt@VeritasPrep GMAT Instructor
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Post Tue Jun 23, 2015 5:54 pm
Inscribing the figure in a circle is the way to go - it works well for any inscribed polygon, not just a triangle.

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amirhakimi Senior | Next Rank: 100 Posts
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Post Thu Oct 31, 2013 7:53 am
Thank you all for those hints. The lines that Brent posted helped a lot.
It was the first time that I post a problem. I didn't know about the tag. So, sorry about that.

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Post Thu Oct 31, 2013 7:39 am
amirhakimi wrote:
In the figure above, triangle ABC is equilateral, and point P is equidistant from vertices A, B, and C. If triangle ABC is rotated clockwise about point P, what is the minimum number of degrees the triangle must be rotated so that point B will be in the position where point A is now?
(A) 60
(B) 120
(C) 180
(D) 240
(E) 270

It may help to add some lines to the diagram.

First add lines from the center to the 3 vertices.
ttp://postimg.org/image/e412m8uq9/" target="_blank">
Aside, we know that each angle is 120º since all three (equivalent) angles must add to 360.º

Then draw a circle so that the triangles vertices are on the circle.
ttp://postimg.org/image/4m5wzxyfl/" target="_blank">

From here, we can see that . ..
ttp://postimg.org/image/npun2jlw1/" target="_blank">
. . . the triangle must be rotated clockwise 240º in order for point B to be in the position where point A is now.


Cheers,
Brent

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